Question

In the following transformation, the number of \( \alpha \) and \( \beta \) particles emitted, respectively, are \[ \begin{matrix} ^{223}_{88}Ra & \xrightarrow{\text{}} & ^{207}_{82}Pb \end{matrix} \]

• A. 4, 2
• B. 4, 3
• C. 2, 4
• D. 3, 4

Hint:

In nuclear decay, each \( \alpha \)-decay decreases the atomic number by 2 and the mass number by 4, while each \( \beta \)-decay increases the atomic number by 1 without changing the mass number.

Answer 1

The Correct Option is A

Step 1: Understanding the nuclear transformation.
This is a nuclear decay reaction where radium-223 decays to lead-207. During this process, the nucleus of radium-223 undergoes a series of decays.
- Each \( \alpha \)-decay decreases the atomic number by 2 and the mass number by 4.
- Each \( \beta \)-decay increases the atomic number by 1 without affecting the mass number.
Step 2: Breakdown of the transformation.
Start with radium-223 (\( ^{223}_{88} \)Ra), and observe the changes:
- For each \( \alpha \)-decay, the atomic number decreases by 2 and the mass number decreases by 4.
- After 4 \( \alpha \)-decays, the atomic number becomes 82 and the mass number becomes 207, which matches lead-207 (\( ^{207}_{82} \)Pb).
- After 4 \( \alpha \)-decays, 2 \( \beta \)-decays are needed to balance the atomic number from 82 to 82 (no change in mass number during \( \beta \)-decay).
Step 3: Conclusion.
The number of \( \alpha \)-particles emitted is 4, and the number of \( \beta \)-particles emitted is 2.
Final Answer: \[ \boxed{4, 2} \]